if 30,72 and x are three integers, such that the product of any two of them is divisible by the third ,then the least value of x is...?

30, 72, and XThere must exist positive integers A, B, and C such that 30·72 = XA 30X = 72B 72X = 30CSimplifying:2160 = XA 5 = 12B 12X = 5C Solving each for X:X = 2160 /AX = 12 B/ 5X = 5 C /12Set the last two right sides equal, since both are equal to X12 *B/ 5 = 5 * C / 12144B = 25CThe smallest B and C can be are B = 25 and C = 144Substituting inX = 12B/ 5 = 12*25/ 5 = 60 X = 5C / 12 = 5 * 144/ 12 = 60That works because30, 72, and 6030*72 = 2160 = 60*3630*60 = 1800 = 73*2560*72 = 4320 = 30*144So the answer is X = 60​